A hotel elevator ascends 200 m with a maximum speed of

5.0 m/s. Its acceleration and deceleration both have a magnitude of 1.0 m/s2.

a. How far does the elevator move while accelerating to full

speed from rest?

b. How long does it take to make the complete trip from bottom to top?

The elevator has a final speed of $v=5m/s$

Initial speed, $u=0$

The magnitude of acceleration,$a=1.0m/s^2$

The elevator moves at a maximum distance,

$$\begin{gathered} d=\frac{v^2-0}{2a} \\ d=\frac{{\left(5m/s\right)}^2}{2\left(1m/s^2\right)} \\ d=12.5m \end{gathered}$$

The elevator moves in two phases- at a constant speed of $v=5m/s$and with an acceleration and deacceleration of magnitude of $1m/s^2$

So the time consumed in moving up and down are the same.

When the elevator is in acceleration, it covers $\left(12.5\times 2\right)m=25m$while moving from bottom to top.

Thus, at a constant speed of$v$,the elevator moves,$d_1=\left(200-25\right)m=175m$

The time taken by the elevator to move upwards and downwards with an acceleration is,

$$\begin{gathered} t=2\times \left(\frac{v-u}{a}\right)\left[Takenv=u+at\right] \\ t=2\times \left(\frac{5-0}{1}\right)s \\ t=10s \end{gathered}$$

Time consumed by the elevator to move with a constant speed is,

$$\begin{gathered} t\text{'}=\frac{175m}{5m/s} \\ t\text{'}=35s \end{gathered}$$

Hence, total time consumed in moving upwards and downwards is,

$$\begin{gathered} T=t+t\text{'} \\ T=10s+35s \\ T=45s \end{gathered}$$